#Restriction – deep mind

In this post, we will introduce the concept of an outer measure, and we will also illustrate the connection to the Lebesgue measure.

Apart from the literature [1] to [6] listed at the end of this post, we also refer to the lecture 245A notes of T. Tao.

The Lebesgue measure is the standard measure that is used to assign real numbers to measurable subsets of the finite-dimensional Euclidean space. One way to construct the Lebesgue measure is to use so-called outer measures. Note that these outer measures also play an important role in dealing with uncertainty via capacities, for instance.

Definition 1.1 (Outer Measure)
Given a set $X$ , let $2^X$ denote the power set. An outer measure on $X$ is a set function $\mu^*: 2^X \rightarrow [0,\infty]$ such that
(i) $\mu(\emptyset)=0$ ;
(ii) $A \subseteq B \Rightarrow \mu^*(A)\leq \mu^*(B)$ ;
(iii) For $A_1, A_2, \ldots \subseteq X$ the following inequality holds:

(1) $\begin{align*} \mu^*\left(\bigcup_{i=1}^{\infty}{A_i}\right) \leq \sum_{i=1}^{\infty}{\mu^*(A_i)}. \end{align*}$

$\square$

Property (ii) is called monotonicity and (iii) is called subadditivity. Note that there are other equivalent definitions that are used to define an outer measure. Outer measures are also known as exterior measures.

Apparently, an outer measure is in general not a measure and the domain is the power set and not the Borel sets $\mathcal{B}(X)$ .

An outer measure only takes positive values since $\emptyset \subset A$ and $\mu^*(\emptyset) \leq \mu^*(A)$ for all sets $A$ . If we furthermore apply (i) of Definition 1.1 to a sequence $A_1, \ldots, A_n, A_{n+1}=\emptyset, \ldots$ , we can imply the finite subadditivity property:

(2) $\begin{align*} \mu^*\left(\bigcup_{i=1}^{n}{A_i}\right) \leq \sum_{i=1}^{n}{\mu^*(A_i)}. \end{align*}$

Let us consider two simple examples. The third example is more complex and will prepare the intuition for the upcoming theorem.

Example 1.1 (Outer Measure)
(a) Consider the set function $\mu^*: 2^{X} \rightarrow \{0,1\}$ defined by $\mu^*(\emptyset)=0$ and $\mu^*(A)=1$ for $\emptyset \neq A\subseteq X$ . The function $\mu^*$ is an outer measure but not a measure. Note that this function does not fulfill the $\sigma$ -additivity.

(b) Consider the set function $\mu^*: 2^{X} \rightarrow \mathbb{N} \cup \{\infty\}$ defined by $\mu^*(A)=|A|$ for finite $A\subseteq X$ and $\mu^*(A)=\infty$ if $A\subseteq X$ is infinite. This set function is also called counting measure, i.e. it is not only an outer measure but also a measure.

(c) Let us now consider the set of all half-open real intervals $\mathcal{F}:=\{[a,b) | \ a,b,\in \mathbb{R}, a\leq b\}$ and define the set function $\lambda([a, b)) = b-a$ that provides the length of the interval. Now, consider

$\begin{align*} & \mu^*:2^{\mathbb{R}} \rightarrow [0, \infty) \text{ defined by } \\ & \mu^*(A):= \inf\{ \sum_{k=1}^{\infty}{\lambda(I_k)} \ | \ I_k\in \mathcal{F}, A \subseteq \bigcup_{k=1}^{\infty}{I_k} \} \end{align*}$

Let us take some time to reflect on this definition of this definition. The real numbers can be illustrated as a continuous line so that any subset of $\mathbb{R}$ can be illustrated as some sort of drawn line that is not necessarily connected.

For instance, the set $A:=A_1 \cup A_2 \cup A_3$ as highlighted in light blue below is not connected and can be approximated by three open intervals $I_1, I_2$ and $I_3$ as illustrated below.

This approximation can become better and better if the half intervals $I_i$ are getting ‘closer’ and ‘closer’ (from the outside) to the sets $A_i$ .

Let us now double-check that the defined function really is an outer measure:

Apparently, $\mu^*$ fulfills (i) of Definition 1.1 since we can pick $\emptyset, \emptyset, \ldots$ as cover.

Let $A \subseteq B$ and let $C:=\bigcup_{k=1}^{\infty}{I_k}$ be an arbitrary cover of $B$ . Then, $A \subseteq B \subseteq C$ , so that every cover of $B$ is also a cover of $A$ . Because of the definition of $\mu^*$ , the desired inequality can be inferred:

$\begin{align*} \mu^*(A) &= \inf \{ \sum_{k=1}^{\infty}{\lambda(I_k)} \ | \ I_k\in \mathcal{F}, A \subseteq \bigcup_{k=1}^{\infty}{I_k} \} \\ \leq \mu^*(B) &= \inf \{ \sum_{k=1}^{\infty}{\lambda(J_k)} \ | \ J_k\in \mathcal{F}, B \subseteq \bigcup_{k=1}^{\infty}{J_k} \}. \end{align*}$

To show the subadditivity, let $\epsilon>0$ and pick $\epsilon_n>0$ such that $\sum_{n\in \mathbb{N}}{\epsilon_n}=\epsilon$ . Then there are intervals $I_{k, n}$ with $\mu^*(A_n) \geq \sum_{k=1}^{\infty}{\mu^*(I_{k, n})}-\epsilon_n$ and $A_n\subseteq \bigcup_{j=1}^{\infty}{I_{k, n}}$ . Then

$\begin{align*} \bigcup_{n\in \mathbb{N}}{A_n} &\subseteq \bigcup_{n\in \mathbb{N}}{\bigcup_{k\in \mathbb{N}}{I_{k, n}}} \\ \Rightarrow & \mu^*\left(\bigcup_{n\in \mathbb{N}}{A_n}\right) \leq \mu^*\left(\bigcup_{n\in \mathbb{N}}{\bigcup_{k\in \mathbb{N}}{I_{k, n}}}\right)\\ &\leq \sum_{k, n\in \mathbb{N}}{\mu^*(I_{k, n})} \\ &= \sum_{n\in \mathbb{N}}{\sum_{k\in \mathbb{N}}{\mu^*(I_{k, n})}} \\ &\leq \sum_{n\in \mathbb{N}}{(\mu^*(A_n)+\epsilon_n)} \\ &= \sum_{n\in \mathbb{N}}{(\mu^*(A_n))} + \epsilon. \end{align*}$

Note that we have applied the monotonicity property, the fact that we approximate the length of $A_n$ by the length of $\bigcup_{j=1}^{\infty}{I_{k, n}}$ and the the assumptions $\mu^*(A_n) \geq \sum_{k=1}^{\infty}{\mu^*(I_{k, n})}-\epsilon_n$ as well as $\sum_{n\in \mathbb{N}}{\epsilon_n}=\epsilon$ .

Since we can choose $\epsilon$ small enough the assertion follows.

$\square$

Example 1.1 (c) illustrates why $\mu^*$ is called an outer measure: it is an outer approximation of an arbitrary set $A$ using a known measure of restricted geometry: intervals and their known lengths are employed to approximate the outer measure of an arbitrary set $A$ .

Generalizing these thoughts lead to the definition of a sequential cover and the subsequent theorem.

Definition 1.2 (Sequential Cover)
A sequential cover of $X$ is a collection $\mathcal{S}$ of subsets of $X$ with the properties that
(i) $\emptyset\in \mathcal{S}$ and
(ii) $\forall A\in 2^X$ there is a countable subcollection $\{I_k\}_{k\in \mathbb{N}} \subseteq \mathcal{S}$ such that $A \subseteq \bigcup_{k\in \mathbb{N}}{I_k}$ .

$\square$

A sequential cover of a set $X$ is therefore a collection of specific subsets $\mathcal{S}$ of $X$ that allows to cover all subsets $A\subset X$ with a countable sub-collection of $\mathcal{S}$ .

In Example 1.1 (c), the set of all half-open intervals $\mathcal{F}$ of $\mathbb{R}$ is an example of a collection of specific subsets of $X=\mathbb{R}$ , which fulfills the requirements of Definition 1.2. These intervals can be used to approximate any other subsets of $X=\mathbb{R}$ .

Note that the definition of compact sets also requires some sort of a finite (sub-)cover. Refer to Definition 4.1 of Fundamentals of Set-Theoretic Topology for the exact definition and further details.

Sequential covers lead to outer measures as follows.

Theorem 1.1 (Extension Theorem )
Assume that $\mathcal{S}$ is a sequential cover of a set $X$ .
If $\lambda: \mathcal{S} \rightarrow [0,\infty)$ is any function for which $\lambda(\emptyset)=0$ , then the function $\mu^*: 2^X \rightarrow [0,\infty]$ defined by

(3) $\begin{align*} \mu^*(A) = \inf \{ \sum_{k=1}^{\infty}{\lambda(I_k)} \ | \ \{I_k\}_{k\in \mathbb{N}} \subseteq \mathcal{S} \\ \text{ and } A \subseteq \bigcup_{k\in\mathbb{N}}{I_k} \} \end{align*}$

in an outer measure on $X$ .

Proof. Note that a similar proof was already given in Example 1.1 (c).

Property (i) of Definition 1.1 is fulfilled since it is required in Theorem 1.1. Let $A \subseteq B$ , then any $\{I_k\}_{k\in \mathbb{N}} \subseteq \mathcal{S}$ that covers $B$ also covers $A$ . Since (3) is taking the infinum, it follows that $\mu^*(A) \leq \mu^*(B)$ . Thus, property (ii) of Definition 1.1 holds true.

All that remains is to verify that $\mu^*$ is countable sub-additive. To this end, suppose that $\{A_k\}_{k\in \mathbb{N}}$ is a sequence of subsets $A_k \subseteq X$ . Since we aim to show that

$\begin{align*} \mu^*\left( \bigcup_{k\in\mathbb{N}}{A_k} \right) \leq \sum_{k\in \mathbb{N}}{\mu^*(A_k)}, \end{align*}$

only the case where the sum converges (i.e. where the terms of the sum are all finite) need to be considered. For this case, suppose that $\epsilon>0$ . For each $k\in\mathbb{N}$ there is a countable family $\{I_{k,n}\}_{n\in\mathbb{N}} \subseteq \mathcal{S}$ such that $A_k \subseteq \bigcup_{n\in \mathbb{N}}{I_{k,n}}$ and

$\begin{align*} \sum_{n\in \mathbb{N}}{\lambda(I_{k,n})} \leq \mu^*(A_k)+ \frac{\epsilon}{2^k}. \end{align*}$

Such a countable family $\{I_{k,n}\}_{n\in\mathbb{N}} \subseteq \mathcal{S}$ must exist since $\mathcal{S}$ is a sequential cover. Thus, $\{I_{k,n}\}_{k,n\in \mathbb{N}}$ forms a countable subcollection of sets from $\mathcal{S}$ that covers $\bigcup_{k\in \mathbb{N}}{A_k}$ and satisfies

$\begin{align*} \mu^* \left( \bigcup_{k\in\mathbb{N}}{A_k} \right) &\leq \sum_{k\in \mathbb{N}}{\sum_{n\in \mathbb{N}}{\lambda(I_{k,n})}} \\ &\leq \sum_{k\in \mathbb{N}}{\mu^*(A_k) + \frac{\epsilon}{2^k}} \\ &\leq \sum_{k\in \mathbb{N}}{\mu^*(A_k)}+\epsilon. \end{align*}$

As $\epsilon>0$ is chosen arbitrarily, $\mu^*$ is indeed countably subadditive.

$\square$

An outer measure is particularly useful since it is easy to construct a measures from it.

Definition 1.3 (Carathéodory Measureable Sets)
Let $\mu^*:2^X \rightarrow [0, \infty]$ be an outer measure and $A \subseteq X$ . The set $A$ is called Carathéodory measurable if

(4) $\begin{align*} \mu^*(Q) = \mu^*(Q \cap A) + \mu^*(Q \cap A^C) \end{align*}$

for all $Q\in 2^X$ . The set of all Carathéodory–measurable sets of $X$ is denoted by $\mathcal{M}(X)$ .

$\square$

Note that some authors require

(5) $\begin{align*} \mu^*(Q) \geq \mu^*(Q \cap A) + \mu^*(Q \cap A^C), \label{def:measurable} \end{align*}$

however, this definition is equivalent to (4) due to requirement (1) of Definition 2.1 of an outer measure. Note that if $\mu^*(Q)$ is not finite, then (4) will be fulfilled anyway.

Let us illustrate the requirement (4) in the following graph with sets $A, A^C, Q\subset X$ .

Recall that outer measures are also used to derive $\sigma$ -additive measures from it. Considering equation (5) and thus the additivity of $\mu^*$ -measurable sets, it is clear that (4) is the first step towards the $\sigma$ -additivity of a measure.

Let us also review the definition of $\mathcal{M}(X)$ , which contains all Carathéodory-measureable sets of $X$ . The set

(6) $\begin{align*} \mathcal{M}(X) = \{ & A \subseteq X \ | \ \mu^*(Q) = \\ & \mu^*(Q \cap A) + \mu^*(Q \cap A^C) \} \end{align*}$

contains all subsets $A$ of $X$ , which can be additively separated by using any other subset $Q\subset X$ . The set $\mathcal{M}$ is going to be important when we want to extend the outer measure to a measure on a $\sigma$ -algebra.

Conclusion 1.1 (Carathéodory Measureable Sets)
Let $\mu^*:2^X \rightarrow [0, \infty]$ an outer measure and $A\subseteq X$ .
(i) If $\mu^*(A)=0$ or $\mu^*(A^C)=0$ , then $A$ is Carathéodory measureable;
(ii) The set $A$ is Carathéodory measureable if for all $Q\subseteq X$ with $\mu^*(Q)<\infty$ the following holds true:

$\begin{align*} \mu^*(Q) \geq \mu^*(Q\cap A)+\mu^*(Q\cap A^C). \end{align*}$

(ii) The set $A$ is Carathéodory measureable if for all $Q\subseteq X$ the following holds true:

$\begin{align*} \mu^*(Q) = \mu^*(Q\cap A)+\mu^*(Q\cap A^C). \end{align*}$

Proof. (i) Let $\mu^*(A)=0$ . Then $\mu^*(Q\cap A)=0$ because of the monotonicity and positivity property of any outer measure (i.e. $Q\cap A \subset A$ ). Thus,

$\begin{align*} \mu^*(Q\cap A)+\mu^*(Q\cap A^C) &=\mu^*(Q\cap A^C) \\ &\leq Q(A). \end{align*}$

Now distinguish the cases when $Q$ is finite and infinite. Now, apply the finite sub-additivity (2). Both cases hold true such that (4) is fulfilled.

(ii) and (iii) follows directly from (i).

$\square$

The next theorem indicates how an outer measure can be extended to a measure.

Theorem 1.2 (Outer Measure, Carathéodory Lemma)
If $\mu^*:2^X \rightarrow [0,\infty]$ is an outer measure, the set $\mathcal{M}(X)$ is a $\sigma$ -algebra and $\mu^*_{|\mathcal{M}(X)}$ is a measure.

Proof. Theorem 4.4 in [1].

$\square$

The last theorem basically tells us that

the Carathéodory-measureable sets is a $\sigma$ -algebra;
an outer measure, restricted to the Carathéodory-measureable sets, is a measure.

Refer to Section 9.3 of [6] or Chapter II – §5 and §6 of [1] for further details.

If we apply this transformation theorem to the Lebesgue measure, we can derive the following quite useful statement.

Theorem 1.3 (Approximation by Open/Closed sets in $\mathbb{R}$ )
A set $A\subseteq \mathbb{R}$ is Lebesgue measurable if and only if, for any $\epsilon>0$ , there exists an open set $O\supseteq A$ , or a closed set $C\subseteq A$ , in $\mathbb{R}$ such that

(7) $\begin{align*} \lambda^*(O \setminus A) &< \epsilon, \quad \text{or}\\ \lambda^*(A\setminus C) &< \epsilon. \end{align*}$

Proof. Theorem 9.4.2 on page 220 in [6].

$\square$

Another good introduction to outer measures is provided by NPTEL founded by the Indian government. It defines outer measure differently and derives the properties (i) to (iii) from Definition 1.1.

Mod-03 Lec-10 / Outer measure and its properties by NPTEL

Literature

[1]

[2]

[3]

[4]

[5]

[6]

Tag: #Restriction

Outer Measures

Literature